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## Hot answers tagged proof

7

Hints: You know the vega of a digital call option formula: $V=-\frac{e^{-r(T-t)}}{\sigma} d_1 n\left(d_2\right)$ Where n is the standard normal density, which is positive. Sigma and exponential are also positive, so the sign of V is down to the sign of $d_1$. Which is negative when: $d_1 <0$ $\ln \frac{S}{X}+\left(r+0.5\sigma^2\right)(T-t)<0$ $S&... 3 overall gamma is second derivative of whole portfolio over underlying. adding any function (such as underlying*constant) which second derivative is 0 does not alter overall second derivative. 2 If you can prove that$\kappa_{\alpha}(X,t)=a_X(\alpha,t^{-1})=t\ln(\frac{E[e^{t^{-1}X}]}{\alpha})$is convex and apply the property of positive homogeneity, then the sub-additivity follows. In original paper, authors show that$\kappa_{\alpha}(X,t)=a_X(\alpha,t^{-1})$is convex in$(X,t)$. Lemma: For fixed$\alpha$, all$\lambda\in[0,1],X,Y\in L_{M^+}$... 1 Recall that for any deterministic function$g,\$ Ito's integral follows a normal distribution: $$\int_0^t g(u) dW_u \sim N\left(0,\int_0^t g^2(u) du\right).$$ Therefore, since $$X_{t+s} - X_t = \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u + \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u,$$ each integral follows a normal distribution and they are ...

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